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$\ln(1+xy) = xy$

When I try to implicitly differentiate this I get

$\frac{1}{1+xy}(y + xy')$ = (y + xy')

At which point the $(y + xy')$ terms cancel out, leaving no $y'$ to solve for.

However, the answer to this is $-\frac{y}{x}$... How do you get this?

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    $\begingroup$ What makes you think you can cancel the $y+xy'$ terms? $\endgroup$ – user1337 Nov 17 '14 at 2:43
  • $\begingroup$ @user1337 Well it is an equation right...? $\endgroup$ – 1110101001 Nov 17 '14 at 2:43
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    $\begingroup$ what if $y+xy'=0$? $\endgroup$ – user1337 Nov 17 '14 at 2:44
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You can only cancel the $y+xy'$ terms if that is not equal to zero, otherwise you're dividing by zero. Assuming that is not equal to zero gets you $$ \frac{1}{1+xy} = 1 $$ so $1+xy=1$, or $xy=0$, which gives you $\log 1=0$ in the original equation. So the solution is based on the notion that for the equation to be non-trivial, $y+xy'=0$.

Incidentally, based on the hint in abel's post, I tried solving the differential equation, and I got $$ {dy \over y} = -{dx \over x} $$ which yields $$ \ln y = -\ln x + c \Rightarrow y=k/x. $$ But, as abel states, the only real solution for the original equation is $k=0$. Wolfram alpha gives a useful graph.

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the only real solution for $\ln(1 + u) = u$ is $u = 0.$ so your equation implies $xy = 0,$ which in turn gives $y = 0$ identically. hence $\frac{dy}{dx} = 0.$

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  • $\begingroup$ This is a very useful contribution. $\endgroup$ – Suzu Hirose Nov 17 '14 at 3:08
  • $\begingroup$ This is how I saw the problem, too. $\endgroup$ – Lubin Nov 17 '14 at 4:20

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